We have had occasion in the preceding pages to return from the second and third powers to their roots. We have shown how this can be done in numeral quantities; it remains to be shown how it may be effected in literal quantities. It is frequently necessary to find the roots of other powers as well as of the second and third.

The power of a literal quantity, we have just seen, is found by multiplying its exponent by the exponent of the power to which it is to be raised.

The second power of a3 is a3 × 2 = a; consequently the second root of a is a

6

= a3.

The third power of am is a3m; hence the third root of a3

In general, the root of a literal quantity may be found by divid ing its exponent by the number expressing the root; that is, by dividing by 2 for the second root, by 3 for the third root, &c. This is the reverse of the method of finding powers.

It was shown above, that any power of a quantity consisting of several factors is the same as the product of the powers of the several factors. From this it follows, that any root of a quantity consisting of several factors is the same as the product of the roots of all the factors.

The third power of a b c is a b3c; the third root of a b c must therefore be a2 b c3.

Numeral coefficients are factors, and in finding powers they are raised to the power; consequently in finding roots, the root of the coefficient must be taken.

The 2nd root of 16 a b2 is 4 a2 b.

When the exponent of a quantity is divisible by the number expressing the degree of the root, the root can be found exactly; but when it is not, the exponent of the root will be a fraction.

The second root of a3 is a3. The second root of a is a3. The third root of a is a. The nth root of a is a. The nth

m

root of am is an.

a3.

1

The root of a fraction is found by taking the root of its numerator and of its denominator. This is evident from the method of finding the powers of fractions.

The root of any quantity may be expressed by enclosing it in a parenthesis or drawing a vinculum over it, and writing a fractional exponent over it, expressive of the root. Thus

The 3d root of 8 a3 b is expressed

The root of a compound quantity may be expressed in the same way.

The 4th root of a2 + 5 a b is expressed

(a2 + 5 a b)1 or a2 + 5 ab$.

When a compound quantity has an exponent, its root may be found in the same manner as that of a simple quantity.

The 3d root of (2 b — a)o is (2 b — a) 3 — (2 b — a)2.

With regard to the signs of roots it may be observed, that all even roots must have the double sign; for since all even powers are necessarily positive, it is impossible to tell whether the power was derived from a positive or negative root, unless something in the conditions of the question shows it. An even root of a negative quantity is impossible. All odd roots will have the same sign as the power.

15. What is the second root of 9 a2 be ?

16. What is the third root of

125 a b c?

17. What is the fifth root of 32 a1o xm r?

20. What is the second root of (2 m — x)o ? 21. What is the 6th root of (3 a + x)m ?

XXXVII. Roots of Compound Quantities.

When a compound quantity is a perfect power, its root may be found; and when it is not a perfect power, its root may be found by approximation, by a method similar to that employed for finding the roots of numeral quantities.

First we may observe, that no quantity consisting of only two terms can be a complete power; for the second power of a binomial consists of three terms; that of a +x, for example, is a2 + 2 a x + x2. The quantity a+b2 is not a complete second power.

Let it be required to find the second root of

9 x* a® + 4 a2 b1 + 12 x2 a1 b2.

The se

The root of this will consist of at least two terms. cond power of the binomial a + b is a2 +2ab+b2. This shows that the quantity must be arranged according to the powers of some letter as in division, for the second power of either term of the root will produce the highest power of the letters in that term.

Arrange the above according to the powers of x.

9 x1 ao + 12 x2 a1 b2 + 4 a2 b*.

The formula a2 + 2 a b + b2 shows that we should find the first term a of the root by taking the root of the first term; the same must be the case in the given example.

The root of 9 x1 ao is 3 x2 a3. Write this in the place of a quotient, and subtract its second power. Then multiply 3 x2 a2 by 2 for a divisor, answering to 2 a of the formula.

9 x* ao + 12 x2 a* b3 + 4 a2 b1 (3x2 a3 +2 ab*

Divide the next term by the divisor. This gives 2 a b2 for the next term of the root. Raise the whole root then to the second power and subtract it. Or, which is the same thing, since the second power of the first term has already been subtracted, write the quantity 2 a b at the right of the divisor as well as in the root. Multiply the whole divisor as it then stands by the last term of the root. This produces the terms corresponding to 2 a b + b2, = b (2 a + b) of the formula. This produces 12 x2 a b2 + 4 a2 b1, which being subtracted, there is no remainder. Consequently the root is 3x a3 + 2 a b2 or 3x2 a3 2 a b'. The second power of both is the same. the double sign had been given to the first term of the root, the second would have had it also, and the positive and negative roots would have been obtained together.

Let it be required to find the 2d root of

If

The process in this case is the same as in the last example. The second term of the root has the sign in consequence of the term 60 a b m2 of the dividend being affected with that sign. If the quantity had been arranged according to the powers of the letter b, thus, 25 b2 60 a b m2 + 36 a2 m2, the root would have been 5 b 6 a m2 instead of 6 a m2 -5 b. Both roots are right, for the second powers of the two quantities are the

same.

b is the same as that of b

α.

The second power a One is the positive and the other the negative root. If the double sign be given to the first term of the root, both results will be produced at the same time in either arrangement.

25 62-60 a b m2 + 36 a2 m2 (±5b6a m2

In dividing-60 a b m2 by the + to, and the to +. the sign. The first value is 6 a m2 - 5 b.

When the quantity whose second root is to be found, consists of more than three terms, it is not the second power of a binomial, but of a quantity consisting of more than two terms. Suppose the root to consist of the three terms m + n +; - p. If we represent the two first terms m+n by l, the expression becomes l+p, the second power of which is

l2 + 21p+ p2.

Developing the second power 12 of the binomial m+n, it becomes m2 + 2 m n + n2. This shows that when the quantity is arranged according to the powers of some letter, the second root of the first term will be the first term m of the root. If m2 be subtracted, and the next term be divided by 2 m, the next term n of the root will be obtained. If the second power of m + n or la be subtracted, the remainder will be 21p+p. If the next term 21 p be divided by 2l equal to twice m +n, the quotient will be p, the third term of the root. The same principle will extend to any number of terms.

It is required to find the second root of

4 a* + 12 a3 x + 13 a2 x2 + 6 α x3 + x*.

Let this be disposed according to the powers of a or of x.